Point Load onto Baseplate/ Ground Bearing Pressure query 2

[sithlord382]

Hi All,

I am working through worked examples on designing base plates (based on Eurocodes) for educational purposes. I have come across what seems to be a computer generated example in which case a 450mm x 450mm x 20dp base plate is subject to a factored point load of 207.68kN.

I am getting stuck on the Load Dispersion figures listed in the snippet below and how this has been derived. I understand that as it is analysing the sufficiency of the existing ground, the unfactored figure of 139.83kN needs to be considered and they have assumed that the allowable bearing capacity is 80kN/m2. However in this example how are they getting to the figure of 74.54kN/m2 for the ground bearing pressure?

I am under the impression (and maybe way off) that ground bearing pressure = Unfactored Force/ Area of footing, but this doesn't coincide with what the example has got for its ground bearing pressure.

Could anyone please assist in explaining this to a 'layman'.

Much Appreciated in advance.

 

[DoubleStud]

It starts as 450mm X 450mm area, then it gets larger when it goes through the concrete slab thickness (30 degree), then it increases again through the hardcore thickness (45 degree). You end up having 202500 mm^2 bearing area.

 

[sithlord382]

Thanks Double Stud for the quick response. Surely 202500 mm^2 is the area of the base plate (450mm x 450mm) alone, before the load disperses into the concrete slab and hardcore?

How does that then result in the figure of 74.54kN/m^2?

 

[DoubleStud]

my bad sith, I didnt check the math and I dont work in mm. So you start with 450 mm, then draw line 30 degree for 150 mm thickness on each side. Then draw line again at 45 degree for 150 mm. That will be your final square dimension. Sorry I was just assuming things on the excel output.

 

[DoubleStud]

sith, I just drew it on autocad and I got 1269.6 mm X 1269.6 mm and give you 86.7 KN/m^2. Not the same as the spreadsheet. Not sure what is going on.

 

[DoubleStud]

Sorry hard to read. My autocad is set up for inches but I got 1269.615 mm

 

[Enable]

They're assuming the 45 degree load goes through the 50mm sand layer as well.

Total per side = 450mm base + 2(260mm for concrete + 200mm for sand & hard) = 1370
Area = 1.37* 1.37 = 1.88m2 ~ 2m2 (rounded)
UDL = 139kN / 1.88m2 ~ 74kN/m2

 

[sithlord382]

Double Stud, could I assume that they have also taken into consideration the 50mm sand the is below the 150mm hardcore? I just use the same principle and used 45deg for the load dispersion into the sand, which gives me an overall base area of 1.369m x 1.369m. That then gets me to the figure of 74.5kN/m^2.

 

[sithlord382]

Double Stud and Enable, honestly guys thank you so much... I have been struggling with this, pulling my hair out, over the last few days. You guys are legends!!

 

[DoubleStud]

Ahh, yes that's it. Very confusing. Why bother separating it? Then there should be a value of dispersion angle for sand.